The lion and the tamer, Part 2

Question

In a previous puzzle we asked whether a lion can catch his tamer if both are contained by a circular cage and ran at the same top speed. Considering the radius of the cage to be r, the lion begins at the center and the tamer at some point on the edge of the cage.

We now remove the cage.

Assume the tamer begins running in a direction perpendicular to the direction of the lion and continues a straight-line course. Further, assume the lion follows a "curve of pursuit." That is, the lion's bearing is always in the direction of the tamer. After a long time, their two paths will become collinear, and the lion will trail the tamer by a certain distance. What is that distance. Again, consider the lion and tamer as points.

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9 answers to this question

I'm embarrassed to admit that I don't know enough math to develop a closed form for the distance.
So, simulating
* the Tamer starting at ( 0,0 ),
* the Lion starting at (0,1 )
* the Tamer moving along X-axis in increments of .01

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No embarrassment needed. It's not simple math. you can Google "pursuit curves" to get an idea. Simulations are the best answer in a lot of cases.

Here's something to think about. If the tamer runs straight away along their joining line, he (obviously) keeps his full distance from the lion. Now we know that if he runs at right angles to his vector from the lion, the separation shrinks to 1/2. This suggests that if he runs at some acute angle to his lion vector (partially towards the lion) the distance would shrink to exactly zero. I don't know, and now I wonder, what angle that would be. Possibly 45 degrees?

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These partial differential equations are beyond me at the moment. But for the follow-on question, I'll posit the following:

Spoiler

Suppose the lion is at some arbitrary angle theta from the tamer's path, which the tamer follows in a straight line without deviation. You can draw this situation by putting the tamer at point (0,0) and running straight up along the y-axis, and putting the lion at some point (x,1) where x is a positive number, so the lion is in the upper right part of the x-y plane.

What happens after some small amount of time dt? The tamer moves up for some distance dt and the lion moves toward the tamer over the same distance. The total distance between the lion and the tamer is reduced, but also notice that, since the lion moves toward the tamer while the tamer moves upward along the y-axis, that means that the new positions after time dt must have the lion at a new angle from the tamer that is greater than the initial angle theta.

The implication of this is that, if the lion could catch the tamer at some angle theta, then the lion would have to also be able to catch the tamer if the starting position were at [angle theta plus a little bit]. Since we know that the lion can't catch the tamer at angle 90 degrees, you can do a proof by induction (or some other similar sort of hand-waving) to convincingly prove beyond all shadow of a doubt that the lion could never, ever, ever catch the tamer unless the tamer were running directly at the lion.

A while back, on another computer whose drive crashed, I simulated the 45-degree case (of a dog chasing a fox, I think, but that doesn't matter) and I don't think the fox was caught. I don't recall finding an angle where there was a capture. I think I'll re-write the program. It's not hard to simulate the chase, and you can get arbitrarily close to the exact solution by decreasing the time increment.

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OK, I searched images the site has stored from previous postings. I found this one, showing a (blue) fox starting from the origin and following various radial paths. A (green) dog starts from (-1, 0) and traces the (green) pursuit paths. The figure does not seem to indicate whether there was a capture.

Note this is not the original lion vs tamer situation -- the red circle is not a cage. In fact, it may have been a modified ogre-maiden situation where the red circle denotes a successful escape for the fox.

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I ran some simulations and it seems the answer is r/2. I tried to put together the diff. equations system and there is no way in the world I can solve such system. How weird that the the solution seems to be so simple...

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The more I tried to simulate Plasmid's argument, the more powerful it felt. While T is moving North, it doesn't matter how close L is to T, or how tiny the angle theta off vertical, with a sufficiently small stepsize, L will waste a portion of each step moving South, widening the angle theta, until eventually | T - L | > stepsize.